Consider the slab shown in Figure. Suppose that$\mathit{L}\mathbf{=}\mathbf{}\mathbf{25}\mathbf{.0}\mathbf{\text{cm}}\mathbf{,}\mathbf{}\mathit{A}\mathbf{=}\mathbf{90}\mathbf{.0}\mathbf{}{\mathbf{\text{cm}}}^{\mathbf{\text{2}}}$, and the material is copper. If${\mathit{T}}_{\mathbf{H}}\mathbf{=}\mathbf{125}\mathbf{°}\mathbf{\text{C}}\mathbf{,}\mathbf{}{\mathit{T}}_{\mathbf{C}}\mathbf{=}\mathbf{}\mathbf{10}\mathbf{.0}\mathbf{°}\mathbf{\text{C}}$, and a steady state is reached, find the conduction rate through the slab.

i) Length$L=25.0\text{cmor0.25m}$

ii) Area$A=90{\text{cm}}^{\text{2}}\text{or}90\times {10}^{-4}{\text{m}}^{\text{2}}$

iii) Temperature$T_C={10}^0\text{C}$

iv) Temperature$T_H={125}^0\text{C}$

v) Thermal conductivity of copper$k=401\text{W/m.K}$

The heat transfer from a hotter body to a colder body is defined as the process of conduction and this heat transfer is known as thermal conduction. The difference rate at which the heat transfer from the hotter end to the other colder end of the system gives us the equation to calculate the rate of conduction which is its power radiation value. The conduction rate is defined as the amount of energy transferred per unit of time.

Formula:

The rate of energy conduction, $P_{cond}=\frac{kA(T_H-T_C)}{L}$ …(i)

Where$k$ is the thermal conductivity of copper,$A$ is the cross-sectional area (in a plane perpendicular to the flow), and$L$ is the distance along the direction of flow between the points where the temperature is data-custom-editor="chemistry" $T_H$and .$T_C$

Using equation (i), the rate of conduction through the slab is given as:

$\begin{array}{c}{P}_{cond}=\frac{\left(\text{401W/m.K}\right)(90\times {10}^{-4}{\text{m}}^{\text{2}})({125}^0\text{C}-{10.0}^0\text{C})}{0.250\text{m}}\\ =1.66\times {10}^3\text{J/s}\end{array}$

Hence, the rate of conduction is$1.66\times {10}^3\text{J/s}$